JEE Main 2022 · 25 July · evening · Single correct · +4 / −1
Light wave travelling in air along x-direction is given by E_y = 540 sin π × 10⁴(x - ct) Vm⁻¹. Then, the peak value of magnetic field of wave will be (Given c = 3 × 10⁸ ms⁻¹)
- (a) 18 × 10⁻⁷ T
- (b) 54 × 10⁻⁷ T
- (c) 54 × 10⁻⁸ T
- (d) 18 × 10⁻⁸ T
Answer and solution
Answer (a) 18 × 10⁻⁷ T
Solution
Physics sir · 4 steps
Read the peak E straight off the sine's amplitude, then divide by c to get the peak B.
- Read off E0. The number in front of the sine is the amplitude, so E₀ = 540 V m⁻¹. The π × 10⁴ inside only sets the wavelength and frequency, so it does not affect the peak values.
- Link E and B. In an electromagnetic wave travelling in air, the electric and magnetic field amplitudes are tied together by the speed of light.
B₀ = E₀/c = 540/(3 × 10⁸) T
- Simplify. 540/3 = 180, so B₀ = 180 × 10⁻⁸ T = 1.8 × 10⁻⁶ T. Written to match the options, this is 18 × 10⁻⁷ T.
- So. The peak magnetic field is 18 × 10⁻⁷ T, which is option (a).
The idea to keep. For any EM wave in vacuum or air, E₀ = cB₀. Take E₀ from the amplitude of the wave equation and ignore what sits inside the sine.
Answer from the NTA final answer key. Solution by PrepWiser.